]]>By: Steven Millerhttp://mathriddles.williams.edu/?p=107&cpage=1#comment-4035
Thu, 26 Jan 2012 16:23:12 +0000http://mathriddles.williams.edu/?p=107#comment-4035ok
]]>By: Geoff Mosshttp://mathriddles.williams.edu/?p=107&cpage=1#comment-4034
Thu, 26 Jan 2012 16:10:22 +0000http://mathriddles.williams.edu/?p=107#comment-4034Ignore this past post, I think I just proved it wrong. Will update after more work.
]]>By: Geoff Mosshttp://mathriddles.williams.edu/?p=107&cpage=1#comment-4033
Thu, 26 Jan 2012 16:07:31 +0000http://mathriddles.williams.edu/?p=107#comment-4033This one is challenging. I’m not 100% certain on the solution, but I’m pretty sure so far. Each general will b given a unique combination. In no particular order, they will be given: a*b, b*c, a*c, a/b, a/c, b/c, a*b/c, a*c/b, b*c/a, and a*b*c. As far as I can tell, no two generals can figure out the code (in some cases, they can solve for a single variable, for example a*b can divide his number by a*b/c, which will give them c, but not a or b), and I believe any combination of 3 should give them all three variables with some basic manipulations.

Not sure how to approach the general case yet, however.

]]>By: Steven Millerhttp://mathriddles.williams.edu/?p=107&cpage=1#comment-3960
Tue, 17 Jan 2012 02:17:33 +0000http://mathriddles.williams.edu/?p=107#comment-3960Sure. This is a fun one, and shows that quadratic equations CAN be fun (ok, I may have a non-standard definition of fun….)
]]>By: Sara Lublashttp://mathriddles.williams.edu/?p=107&cpage=1#comment-3958
Mon, 16 Jan 2012 19:24:31 +0000http://mathriddles.williams.edu/?p=107#comment-3958Hiya, could you please send me the answer to storms999@gmail.com
Really struggling on this one, and really wanna know how to do this! 😀
Thanks a lot
]]>By: Steven Millerhttp://mathriddles.williams.edu/?p=107&cpage=1#comment-2689
Sat, 17 Dec 2011 14:23:44 +0000http://mathriddles.williams.edu/?p=107#comment-2689email to you bounced — can you try emailing me at sjm1@williams.edu
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